\subsection{Root sums and Schur degrees}
\label{fac:degree}

Define the two signed short centers by
\begin{equation}\label{fac:centers}
 m=\frac{b+c}{2},\qquad d=\frac{b-c}{2}.
\end{equation}
For \(r\geq0\) and \(t\in\mathbb R\), write \(K_r(t)=\max(0,r-|t|)\), as in Section~\ref{sec:spectral}, and use \(u_+=\max(u,0)\) for \(u\in\mathbb R\). For a list \(R=(R_1,\ldots,R_N)\) of real exponents, let
\begin{equation}\label{fac:root-degree}
 A_R=\sum_{i<j}\bigl(K_1(R_i-R_j)+K_1(R_i+R_j)\bigr)
       +\sum_i\bigl(K_{|m|}(R_i)+K_{|d|}(R_i)\bigr).
\end{equation}
Here \(1\leq i<j\leq N\) and \(1\leq i\leq N\) in the respective sums. This is \eqref{eq:root-degree} with middle weight one. In the recursions the list may be empty, in which case the sum is zero.

Use the bipartition labels for the simple finite type \(B\) modules, with the upper partition \(\alpha\) associated with the trivial coordinate character and the lower partition \(\beta\) with the nontrivial one. For endpoint weight \(t\geq0\), let \(a_t(\alpha,\beta)\) be minus half the smallest Schur exponent. The finite spectral degree is \(a_t\) at the label twisted by the full sign, the one-dimensional character taking every simple reflection to \(-1\). Indeed \(v\mapsto v^{-1}\), \(N_s\mapsto -N_s\) is semilinear, preserves the symmetrizing trace in the sense of that parameter change, and specializes to the sign twist. We calculate with these twisted labels. A partition \(\alpha=(\alpha_i)_{i\geq1}\) has weakly decreasing nonnegative integer parts and a zero tail. Put \(|\alpha|=\sum_i\alpha_i\) and \(n(\alpha)=\sum_i(i-1)\alpha_i\). The charged-symbol formula is
\begin{equation}\label{fac:symbol}
 a_t(\alpha,\beta)=n(\alpha)+n(\beta)+
 \sum_{i,j\geq 1}
 \left(\min(\alpha_i-i+t,\beta_j-j)-\min(-i+t,-j)\right).
\end{equation}
For \(t\geq0\), specialize \cite[Section~4.3]{ChlouverakiJacon} with ordinary parameter \(q=v^{2a}\) and eigenvalues \(q^t,-1\) for the unnormalized endpoint generator. In the shifted symbol, subtract the empty-symbol pair contributions and use common padding in the two rows. Any extra entries from the longer row lie below all nonzero parts of the opposite row for sufficiently large padding and cancel against the empty symbol. Within either row the minimum for indices \(i<j\) is the entry at \(j\), giving \(n(\alpha)\) and \(n(\beta)\). The cross terms then give the displayed sum, in which only finitely many summands are nonzero. For real \(t\geq0\), the valuation is given by the same continuous piecewise-linear formula, since each cross factor of the Schur product is a sum of two positive monomials. For signed \(t\) extend \(a_t\) by \eqref{fac:symbol}. Interchanging the two rows gives \(a_{-t}(\beta,\alpha)=a_t(\alpha,\beta)\). For example, the rank-one trivial type has Schur element \(1+q^t\), so its normalized spectral degree is \(t=a_t(\varnothing,(1))\).

\paragraph{Deletion of one row.}
Suppose a row of length \(g\geq1\) has exponents \(m,m+1,\ldots,m+g-1\), and \(m+g-1\) is their largest absolute exponent and dominates the absolute values of all remaining exponents. Remove this row and replace \(b,c\) by \(b-1,c-1\). The remaining exponents stay fixed. For any remaining exponent \(t\), the sum of the difference and sum root contributions from the deleted row is
\[
 H(m-t)+H(m+t),\qquad H(u)=(1-u)_+-(-u)_+ .
\]
Indeed the consecutive functions \(K_1(t-m-j)\), \(0\leq j<g\), sum to one between the first and last centers and rise linearly from zero to one on \([m-1,m]\). For \(t\leq m+g-1\) this sum is \(H(m-t)\). The same calculation at \(-t\) gives the sum-root contribution. The identity
\[
 H(m-t)+H(m+t)+K_{|m|}(t)-K_{|m-1|}(t)=1
\]
follows from \((x+t)_++(x-t)_+=x+\max(|x|,|t|)\). Thus the root degree of the full list is the new root degree of the remaining list plus the number of remaining exponents and the root degree of the removed row.

The root degree of the removed row is
\begin{equation}\label{fac:strip-degree}
 \sum_{j=0}^{g-1}\bigl((-b-j)_++(-c-j)_+\bigr).
\end{equation}
To check this, write \(r_j=m+j\). The difference-root terms vanish, and
\[
 (-r_j-d)_++(-r_j+d)_+=K_{|d|}(r_j)+2(-r_j)_+.
\]
It suffices to show that the sum-root terms and the \(|m|\)-potential add to \(2\sum_j(-r_j)_+\). For \(m\geq0\) both are zero. Define \(F(u)=\sum_{j\geq0}(u-j)_+\) for any real \(u\). For \(m<0\), put \(t=-m\). Since \(g-1\geq 2t\), the \(|m|\)-potential is \(F(2t)-2F(t)\). Write \(t=h+\theta\) for \(h=\lfloor t\rfloor\), \(0\leq\theta<1\). For \(j<\ell\), only the pairs with \(j+\ell=\lfloor 2t\rfloor\) or \(\lfloor 2t\rfloor+1\) can contribute to \(K_1(j+\ell-2t)\). There are \(\lfloor (u+1)/2\rfloor\) such pairs with \(j+\ell=u\), and every pair of positive contribution has \(\ell<g\). Hence their sum is \(h+2\theta\) if \(\theta\leq1/2\), and \(h+1\) otherwise. By the finite progression formula for \(F\) this is \(4F(t)-F(2t)\), as required.

For \eqref{fac:symbol}, prepend an upper row of integer length \(x\geq0\) to \(\alpha\), writing \((x,\alpha)\) for the resulting partition. Suppose \(x\geq \alpha_1\) and \(x+t-1\geq\beta_j-j\) whenever \(\beta_j>0\). Its effect is
\begin{equation}\label{fac:single-schur}
 a_t((x,\alpha),\beta)=a_{t-1}(\alpha,\beta)+|\alpha|+|\beta|
                       +\sum_{j=0}^{x-1}(-t-j)_+ .
\end{equation}
The surviving upper entries have charge \(t-1\). The new row adds \(|\alpha|\) to the within-color term, and its cross term is \(|\beta|\) plus the difference of its empty-opposite-row contributions. Increasing \(x\) by one changes the latter difference by \((-t-x)_+\), by telescoping the minima against the entries \(-j\), \(j\geq1\). This also proves \eqref{fac:single-schur} for \(x=0\) when the stated inequalities hold, by using a leading zero in the padded symbol.

A split into upper rows \(x,g-x\) at the left and right endpoints realizes \eqref{fac:strip-degree} provided
\begin{equation}\label{fac:strip-box}
 \min(g,\max(0,\lceil-c\rceil))\leq x,\qquad
 \min(g,\max(0,\lceil-b\rceil))\leq g-x .
\end{equation}
These bounds keep all positive summands in the two finite progression sums truncated at \(x\) and \(g-x\), respectively. They apply to the row of positive phase. For a column of positive phase, change both endpoint signs and interchange both colors. For a row of negative phase, change the left endpoint sign and interchange its two colors. For a column of negative phase, change the right endpoint sign and interchange its two colors. These changes leave \eqref{fac:root-degree} and \eqref{fac:symbol} invariant after the corresponding relabeling.

\paragraph{Deletion of a balanced string.}
For an integer \(k\geq1\), a balanced string is the list \(R_k=((k-1)/2-j)_{0\leq j<k}\). Let
\[
 \begin{gathered}
 C(k,u)=\max(0,\min(k,\lfloor(k+1)/2-u\rfloor)),\\
 e=C(k,m),\quad f=k-e,\quad E=C(k,d),\quad F_0=k-E .
 \end{gathered}
\]
In the generic region \(b,c,b+c,b-c\notin\mathbb Z\), \(C(k,u)\) is the clipped nearest integer to \(k/2-u\) at each of the signed centers \(u=\pm m,\pm d\), and \(C(k,-u)=k-C(k,u)\). We choose four rows, with left upper and lower lengths \(x,y\), and right upper and lower lengths \(e-x,f-y\), subject to
\begin{equation}\label{fac:packet-box}
 0\leq x\leq e,\quad 0\leq y\leq f,\qquad x-y=e-E .
\end{equation}
Their upper-upper, lower-lower, lower-upper and upper-lower sums across the two endpoints are \(e,f,E,F_0\). In particular the same four constraints accommodate a negative-phase string by exchanging the two left colors.

Every solution to \eqref{fac:packet-box} has
\begin{equation}\label{fac:packet-energy}
 a_c((x),(y))+a_b((e-x),(f-y))
  =\lfloor k/2\rfloor+\sum_{r\in R_k}\bigl(K_{|m|}(r)+K_{|d|}(r)\bigr).
\end{equation}
Here and below a single part of size zero gives the empty partition. Both sides are invariant under exchanging the endpoints, or negating an endpoint weight and exchanging its colors. These operations permute the four row-sum constraints. Thus assume \(b\geq c\geq0\), so \(m\geq d\geq0\) and \(e\leq E\leq\lfloor k/2\rfloor\). For \(t\geq0\) and \(0\leq z\leq w\), the cancellation-free hook formula \cite[Proposition 4.5]{ChlouverakiJacon} gives
\begin{equation}\label{fac:two-rows}
 a_t((z),(w))=z+F(t)-F(t-w-1)-(t-w+z)_+.
\end{equation}
In fact the merged row partition contributes \(z\) and the upper hooks contribute zero to the negative valuation. The lower generalized hook in column \(j\) is \(w-j+\mathbf 1_{j\leq z}\), \(1\leq j\leq w\). Its contribution is \((t-w+j-\mathbf 1_{j\leq z})_+\), which sums to the last three terms of \eqref{fac:two-rows}.

For a real \(u\), put \(P(u)=(u-\lfloor u\rfloor)(1-u+\lfloor u\rfloor)\). Direct summation gives \(2F(u)=u^2+u+P(u)\) for \(u\geq-1\). Also, for \(|U|,|V|\leq1/2\),
\begin{equation}\label{fac:period}
 \frac{P(U+V)+P(U-V)}2=\max(|U|,|V|)-U^2-V^2.
\end{equation}
The left side is even in each variable and symmetric in the two variables. For \(0\leq V\leq U\leq1/2\), both arguments are in \([0,1]\), so the displayed identity is the quadratic formula for \(P\) on that interval.

Let \(M=m+e-k/2\) and \(D=d+E-k/2\). If \(e>0\), then both \(|M|,|D|<1/2\). In \eqref{fac:two-rows}, the first endpoint has \(z=x,\ w=x+E-e\) and the second has \(z=e-x,\ w=k-E-x\). The two terms \(F(t-w-1)\) vanish because their arguments are \(M-D-x-1\) and \(M+D-(e-x)-1\). The sum is
\[
 e+F(b)+F(c)-(M+D)_+-(M-D)_+
 =\frac{k}{2}+m^2-M^2+d^2-D^2,
\]
Indeed \(b\equiv M+D\) and \(c\equiv M-D\) modulo \(\mathbb Z\). Substituting the quadratic expression for \(F\) and using \(b^2+c^2=2(m^2+d^2)\) gives
\[
 F(b)+F(c)=m^2+d^2+m+\max(|M|,|D|)-M^2-D^2.
\]
Now \((M+D)_++(M-D)_+=M+\max(|M|,|D|)\) and \(e+m-M=k/2\).
If \(e=0<E\), then \(x=0,y=E\), \(M>-1/2\), and \(|D|<1/2\). The two lower rows contribute
\[
 F(c)+F(b)-F(M-D)-F(M+D)
  =\frac{k}{2}+m^2-M^2+d^2-D^2,
\]
since \(c=M-D+E\), \(b=M+D+k-E\), and \(M\pm D>-1\). Subtracting the quadratic expressions for the two pairs eliminates \(P\) by periodicity and gives \(kM+(k-2E)D+\bigl(E(E+1)+(k-E)(k-E+1)\bigr)/2\), which is the right side. Finally, if \(e=E=0\) the contribution is \(F(b)-F(b-k)=kb-k(k-1)/2\), since \(b=m+d>k-1\).

For \(u\geq(k-1)/2\), every string entry satisfies \(|r|\leq u\), so \(\sum_{r\in R_k}K_u(r)=ku-(k^2-\epsilon)/4\), where \(\epsilon\) is the parity of \(k\). For \(0\leq u\leq(k+1)/2\), when \(k\) is odd put \(j=\lfloor u\rfloor\). The sum is \((2j+1)u-j(j+1)\), by pairing the terms at the positive and negative integers around zero. When \(k\) is even put \(j=\lfloor u+1/2\rfloor\). The sum is \(2ju-j^2\), by pairing the terms at the positive and negative half-integers. Any missing term at the upper endpoint \(u=(k+1)/2\) has coefficient zero. In either parity this is \(u^2-\operatorname{dist}(u-k/2,\mathbb Z)^2+\epsilon/4\), where the distance is the minimum absolute difference from an integer. Substitution in the three cases gives \eqref{fac:packet-energy}. The difference-root contributions within \(R_k\) vanish, and each unordered pair of distinct opposite entries contributes one sum-root term. There are \(\lfloor k/2\rfloor\) such pairs, proving the standalone root degree formula.

When a balanced string has largest radius among the remaining absolute exponents, both \(t\) and \(-t\) lie between its extreme centers for each exponent \(t\) left after deletion. Hence both sums \(\sum_{r\in R_k} K_1(r\pm t)\) are one, giving twice the length of the remaining list. The endpoint weights are unchanged. For \eqref{fac:symbol}, prepend an upper and a lower row at one endpoint with remaining partitions \(\alpha,\beta\). If the new rows give partitions and each new symbol entry is at least every positive-part symbol entry in the opposite tail, each new cross term against a positive opposite part increases by that part over the term against zero. The two new cross sums thus contribute \(|\alpha|+|\beta|\) beyond the standalone two-row degree. The within-color terms increase by \(|\alpha|+|\beta|\) and the cross terms among the old rows are unchanged because both rows and their empty counterparts shift by one. The increments from the two endpoints add to twice the length of the remaining list, as required.


\subsection{Splitting the rows between the endpoints}
\label{fac:splitting}

Take two partitions \(\lambda,\mu\), whose box exponents are \(m+j-i\) and \(d+j-i\), respectively, for a box in row \(i\) and column \(j\). Also take a list of balanced strings. At each step delete the object containing an exponent of largest absolute value. For a partition compare its upper-right box exponent with the negative of its lower-left box exponent. Delete the first row for the former extremum and the first column for the latter. For a balanced string delete the entire string. The signs of a strip, in left-right order, are \((+,+)\) for a row of \(\lambda\), \((-,-)\) for a column of \(\lambda\), \((-,+)\) for a row of \(\mu\), and \((+,-)\) for a column of \(\mu\). On deleting a strip with signs \((s,t)\), replace \(c,b\) by \(c-s,b-t\), respectively. The exponent of each undeleted box stays fixed under this reindexing. Deleting a string leaves the parameters fixed. Throughout the generic region the strip extrema have distinct absolute values across the two partitions and cannot tie a string radius. Strings of equal length can be ordered arbitrarily.

The two endpoint bipartitions are constructed by reversing the deletion order. Number the deletions, called events below, in the order they occur. At each endpoint a strip contributes one entry in its assigned color, and a string contributes an entry in each color. Count entries of each color in event order starting at one, including entries of length zero. Write \(r(i,s)\) for the resulting index of an entry of color \(s\in\{+,-\}\) at event \(i\). For entries of colors \(s,t\) at events \(i<j\) and initial endpoint weight \(u\), define
\[
 \delta((i,s),(j,t))=
 \begin{cases}
  0,&s=t,\\
  \lceil r(i,s)-r(j,t)-u(\mathbf 1_{s=+}-\mathbf 1_{t=+})\rceil,&s\ne t.
 \end{cases}
\]
Let \(z(i,s)\) be the length of the entry. We impose
\begin{equation}\label{fac:ordering}
 z(i,s)\geq z(j,t)+\delta((i,s),(j,t))
 \quad\text{when }i<j\text{ and }z(j,t)>0.
\end{equation}
These conditions give partition order in each color and the dominance of each newly prepended entry over the later positive entries of the opposite color. They give exactly the hypotheses of the degree recursions in Section~\ref{fac:degree}. For a strip of length \(g\) with signs \((s,t)\), the two lengths sum to \(g\), and the standalone lower bounds are \(\min(g,\max(0,\lceil-sc\rceil))\) and \(\min(g,\max(0,\lceil-tb\rceil))\), at its current parameters. For a string use \eqref{fac:packet-box} at its current parameters.

\begin{lemma}\label{fac:extension}
Under \(b,c,b+c,b-c\notin\mathbb Z\), every choice of lengths on a suffix of the deletion list satisfying the standalone bounds and \eqref{fac:ordering} extends through the preceding event.
\end{lemma}

\begin{proof}
A positive row of either color in the suffix forces its first row of that color to be positive. Only that first row can give the strongest constraint on a prepended row, since subsequent lengths decrease and the opposite-color shifts in \eqref{fac:ordering} also decrease. Each relevant first row lies at or before the first string event in the suffix, since a string has both colors. In particular, if two such rows come from distinct events, the earlier event is a strip.

At current endpoint weight \(u\), the new row of sign \(s\) is bounded below by the maximum of zero, the first same-color length, and the first opposite-color length plus \(\lceil-su\rceil\) for a strip. For a string the last shift is \(\lceil-su-1\rceil\), because the string also adds a row of the opposite color. Terms from empty colors are omitted. These formulas follow by counting earlier upper and lower entries in \eqref{fac:ordering}. Equivalently, after removing any initial sequence of events and updating the endpoint weight by their signs, the index differences give the same shifts on the remaining events.

We use the inequality
\begin{equation}\label{fac:triangle}
 \delta(i,k)\leq\delta(i,j)+\delta(j,k)
\end{equation}
at one endpoint when \(i<j<k\) and \(k\) is the first entry of its color after \(i\). The entry signs are implicit in this notation. If the signs of \(i,j\) agree, the index at \(i\) is at most that at \(j\), and if the signs of \(j,k\) agree, the index at \(k\) is at least that at \(j\). These observations give \eqref{fac:triangle} in both cases. In the alternating case, the two opposite-color shifts sum to \(r(i,s)-r(k,s)+1=0\), since the endpoint weight is nonintegral and \(k\) is first of its color. This proves \eqref{fac:triangle}, including when \(i\) is one of the entries of a string.

Independent sign changes of the two endpoint weights with the corresponding color exchanges preserve the inequalities. More explicitly, multiplying the left and right signs and weights by \(l,r\in\{1,-1\}\) relabels the phases by multiplication by \(lr\) and transposes both residual partitions if \(r=-1\). All box exponents are multiplied by \(r\). The signed center \((tb+sc)/2\) associated with a pair of endpoint signs \((s,t)\) is unchanged under the transformed pair. The four string capacities are \(C(k,(tb+sc)/2)\), so they are permuted by this operation. Exchanging the endpoints and transposing \(\mu\) is another symmetry. We may therefore take a current strip to have signs \((+,+)\). For the current string it suffices to bound the sum of the requirements for the two upper rows.

\paragraph{Prepending a strip.}
Let the current strip be the first row of \(\lambda\), of length \(g\). Let \(h\) be the height of \(\lambda\), and put \(K=\lceil-c\rceil\), \(J=\lceil-b\rceil\). The radius of the strip is \(M=m+g-1\). Comparing the other extrema with \(M\) bounds \(h\) by \(g+\lfloor b+c\rfloor\), and bounds the width and height of \(\mu\), when nonempty, by \(g-K\) and \(g-J\). Every later strip of signs \((+,+),(-,+),(+,-),(-,-)\) has length at most
\[
 g,\quad g-K,\quad g-J,\quad g+\lfloor b+c\rfloor-1,
\]
respectively. Its two shifts from the current rows add to at most \(0,K,J,K+J\) in those four cases. Thus its length plus its two shifts is at most \(g\), using \(\lfloor b+c\rfloor-\lfloor b\rfloor-\lfloor c\rfloor\leq1\).

Consider two positive tail requirements at different events, one at each endpoint. If the first comes from the left, its event \(j\) is a strip. Write \(k\) for the event of the right requirement. By \eqref{fac:ordering} at the right and \eqref{fac:triangle}, the right requirement is at most the right row of \(j\) plus its shift from the current event. The sum of the two requirements is therefore at most the length of the strip \(j\) plus its two shifts, and hence at most \(g\). Exchanging endpoints gives the same conclusion when the first requirement comes from the right. When both requirements come from one strip, the same length bound applies directly.

If both positive requirements come from one string of size \(k'\) with selected colors \((s,t)\), every intervening strip has signs \((-s,-t)\). Let \(p\) be their number and \(M'=(k'-1)/2<M\) the radius of the later string. Its selected capacity is the clipping to \([0,k']\) of the floor of
\[
 M'+1-\frac{tb+sc}{2}+\frac{s+t}{2}-p .
\]
This follows by applying the strip updates to the two weights. The untruncated floors in the four sign cases are bounded respectively by
\[
 g-p,\quad g-K-p,\quad g-J-p,\quad
 g+\lfloor b+c\rfloor-1-p .
\]
The selected capacity is positive, so clipping can only decrease the floor. Adding the shifts \(K\) or \(J\) for each opposite color gives again a bound of \(g\).

It remains to pair a tail requirement at one endpoint with the standalone requirement at the other. We prove that each retained right requirement \(B\) satisfies \(B\leq g-\max(0,K)\). Swapping endpoints proves the left version. Suppose its source is a strip. If its right sign is \(+\), the preceding strips all have right sign \(-\). Let \(p,q\) count their \((-,-),(+,-)\) signs. The left weight there is \(c-1+p-q\) and the shift on the right is zero. If its right sign is \(-\), let \(p,q\) count the preceding \((+,+),(-,+)\) strips. The left weight there is \(c-1-p+q\) and its right shift is \(J\). In these two cases, the retained positive right row forces its companion left row to be at least the ceiling of minus its left sign times the left weight there. Indeed a ceiling at least the full strip length would force the right row to vanish. Combining this companion bound with the length bounds gives the following two bounds in each case:
\[
\begin{array}{c|cc}
\text{source signs}&\multicolumn{2}{c}{\text{upper bounds for }B}\\ \hline
(+,+)&g-p&g-K-1-q\\
(-,+)&g-K-q&g-p\\
(+,-)&g-q&g-K-1-p\\
(-,-)&g-K-p&g-q
\end{array}
\]
For the first two rows the source lengths are at most \(g-p\) and \(g-K-q\), respectively. For the third and fourth they are at most \(g-J-q\) and \(h-1-p\). In the fourth row use \(h-1+K+J\leq g\). These bounds give the claimed inequality in every sign case.

Suppose the retained right row is from a string. For a right upper row let \(p,q\) count the preceding \((-,-),(+,-)\) strips. The two centers there are \(m-1+p,d+q\). Its length is at most both \(C(k',m-1+p)\) and \(C(k',d+q)\). The radius inequality \(M'<M\) gives upper bounds \(g-p,g-K-q\) for the untruncated floors. For a right lower row let \(p,q\) count the preceding \((+,+),(-,+)\) strips. The centers there are \(m-1-p,d-q\). Its length is at most both \(C(k',-m+1+p)\) and \(C(k',-d+q)\), whose untruncated floors are at most \(g+\lfloor b+c\rfloor-1-p,g-J-q\). These capacities are positive. Adding the shift \(J\) in the lower case gives bounds \(g-K-p,g-q\). Thus every retained right requirement is at most \(g-\max(0,K)\).

Finally the two standalone bounds sum to at most \(g\). If both \(K,J>0\), then \(h\geq1\) implies \(\lfloor b+c\rfloor\geq1-g\) and hence \(K+J\leq g\). If one is nonpositive, its standalone bound is zero. The pairs of standalone and tail bounds are compatible by the unary bounds just proved, and every pair of tail bounds is compatible by the two-target argument. Taking the maximum of the bounds at each endpoint thus gives two nonnegative integers with sum at most \(g\). They admit a split of the strip of length \(g\).

\paragraph{Prepending a balanced string.}
Let \(k\) be its length and use \(e,f,E,F_0\) as in \eqref{fac:packet-box}. Denote the four nonnegative integer requirements from the suffix by \(l_+,l_-,r_+,r_-\), with \(l\) and \(r\) indicating the two endpoints. On the diagonal \(x-y=e-E\), the possible upper-left values are exactly the integers in
\[
 \bigl[\max(l_+,l_-+e-E),\,\min(e-r_+,f-r_-+e-E)\bigr].
\]
Thus existence is equivalent to the four inequalities
\begin{equation}\label{fac:cuts}
 l_++r_+\leq e,\quad l_-+r_-\leq f,\quad
 l_-+r_+\leq E,\quad l_++r_-\leq F_0 .
\end{equation}
The sign changes above reduce them to \(l_++r_+\leq e\).

Assume \(0<e<k\). For a nonempty residual partition at center \(u\), the first box gives \(|u|\leq(k-1)/2\). Its upper-right and lower-left boxes bound its width and height by \(\lfloor(k+1)/2-u\rfloor\) and \(\lfloor(k+1)/2+u\rfloor\). These floors are in \([0,k]\) by the first-box bound. Empty partitions have width and height zero. Thus the bounds are \(e,f\) for \(\lambda\) and \(E,F_0\) for \(\mu\). Let \(E'\) and \(F'\) be the two unclipped floors \(\lfloor (k+1)/2-d\rfloor,\lfloor (k+1)/2+d\rfloor\). The floor difference for two arguments separated by \(c\), respectively \(b\), gives
\[
 e-E'\in\{K-1,K\},\qquad e-F'\in\{J-1,J\},\qquad E'+F'=k.
\]
In particular \(E+K-1\leq e\) when \(E>0\), \(F_0+J-1\leq e\) when \(F_0>0\), and \(f+K+J-2\leq e\). These bounds show that any later strip length plus its two shifts from the current upper rows is at most \(e\). In fact the four length bounds are \(e,E,F_0,f\), and their shift bounds are \(0,K-1,J-1,K+J-2\), respectively. For two positive requirements at distinct events, or from one strip, \eqref{fac:triangle} and the strip argument above now give total at most \(e\). If both come from the same later string, every intervening strip has both colors opposite to its two selected colors. Its corresponding signed center has increased by the number of those strips. Its size has weakly decreased, so its capacity is at most the current capacity for those signs, since \(C(k,u)\) increases in \(k\) and decreases in \(u\). Adding the shifts bounded by \(K-1,J-1\) for the opposite colors gives again a total at most \(e\).

To pair a positive requirement with zero, it suffices by endpoint exchange to bound each left requirement by \(e\). If its source has left color \(+\), let \(p,q\) count the preceding \((-,+),(-,-)\) strips. The centers at its event are \(m+q,d-p\), and the right weight is \(b-p+q\). A \((+,+)\) source has length at most \(e-q\). A \((+,-)\) source has length at most \(F_0-p\), with right companion at least \(\lceil b-p+q\rceil=1-J-p+q\) by the standalone bound and positivity of the left row. Its left row is therefore at most \(F_0+J-1-q\leq e\). A string source has left upper length at most \(C(k',m+q)\leq e\), with \(k'\leq k\). If the source has left color \(-\), let \(p,q\) count the preceding \((+,+),(+,-)\) strips. The centers are now \(m-p,d+q\) and the right weight is again \(b-p+q\). The left shift is \(K-1\). A \((-,+)\) strip has length at most \(E-q\), giving a shifted bound at most \(e\). A \((-,-)\) strip has length at most \(f-p\) and right companion at least \(1-J-p+q\), giving a shifted bound \(f+K+J-2-q\leq e\). A string has left lower length at most \(C(k',d+q)\leq E\). Its shifted bound is at most \(E+K-1\leq e\). This proves \(l_++r_+\leq e\) for \(0<e<k\).

If \(e=0\), then \(m>(k-1)/2\) and \(\lambda\) is empty. The center \(m\) stays fixed. A later row of \(\mu\) at center \(d'\), of length \(g\), satisfies \(d'+g-1\leq(k-1)/2<m\). Its left lower bound is \(\min(g,\max(0,\lceil m-d'\rceil))=g\), so its right upper row vanishes. A column of \(\mu\) similarly has vanishing left upper row. Every later string has zero upper-upper capacity. If a first positive left lower row occurs, all preceding strips are columns of \(\mu\). If there are \(q\) such columns, its center is \(d+q\). The radius bound gives length at most \(\lceil c\rceil-q\) for a strip. For a string this same bound follows from the unclipped floor of \(C(k',d+q)\), using \(k'\leq k\). Hence its requirement on the new left upper row is at most \(\lceil c\rceil+\lceil-c-1\rceil=0\). The right upper requirement also vanishes by endpoint exchange.

If \(e=k\), the simultaneous sign and color reversal of the preceding argument shows that \(\lambda\) is empty and all lower tail rows vanish. The first left upper row, if present, follows only rows of \(\mu\). If there are \(p\) such rows, a column source has length at most the initial height of \(\mu\) minus \(p\), bounded by \(F_0-p\). A string source has left upper length \(C(k',-d+p)\leq F_0\), since its right lower row is zero. Therefore \(l_+\leq F_0\). Endpoint exchange gives \(r_+\leq E\), so \(l_++r_+\leq F_0+E=k\). This finishes all four inequalities in \eqref{fac:cuts} and the proof.
\end{proof}

Starting with the empty suffix, Lemma~\ref{fac:extension} yields two endpoint bipartitions. The degree recursions show that their Schur valuations add to the root degree of the original list. The signs determine how a row or a column of a phase partition is split. For strings, the four capacities give the prescribed row totals in either phase.